1. q lost = q gained

2. Cooling object 7.2 mol of H 2 gas is cooled from 319 K to 299K, how much heat was lost?7.2 mol of H 2 gas is cooled from 319 K to 299K, how much heat was lost? q = 7.2mol (28.8J/molK)(299-319K)q = 7.2mol (28.8J/molK)(299-319K) q = -4147.2 Jq = -4147.2 J q = -4100 Jq = -4100 J 4.1 kJ was lost4.1 kJ was lost negative means the energy was lost instead of gained.negative means the energy was lost instead of gained.

3. Energy is never destroyed!!! 1 st law of thermodynamics1 st law of thermodynamics In a closed system, if you put 2 objects of different temperatures together…In a closed system, if you put 2 objects of different temperatures together… the heat will go from the hotter object to the cooler object.the heat will go from the hotter object to the cooler object. the q lost by one object will be the q gained by the other object.the q lost by one object will be the q gained by the other object. q lost = q gainedq lost = q gained

4. continuing with the H 2 from the earlier problem The hydrogen was put in 9.179 moles of a substance at 274 K and it heated to 280 K, what was the substance?The hydrogen was put in 9.179 moles of a substance at 274 K and it heated to 280 K, what was the substance? q lost = 4147.2 Jq lost = 4147.2 J 4147.2 = 9.179 mol C (280-274K)4147.2 = 9.179 mol C (280-274K) C = 75.3 J/K molC = 75.3 J/K mol water (liquid)water (liquid)

5. Another problem.778 mol of tungsten at 68 o C is dropped in water at 25 o C, the system comes to equilibrium (both temperatures are equal) at 35 o C. How much water was present?.778 mol of tungsten at 68 o C is dropped in water at 25 o C, the system comes to equilibrium (both temperatures are equal) at 35 o C. How much water was present? q lost =.778 mol(24.2)(-33)q lost =.778 mol(24.2)(-33) = 621.3108 = n (75.3)(10)= 621.3108 = n (75.3)(10) n =.83 mol H 2 On =.83 mol H 2 O

6. A problem You add 14.2 g of a metal at 98.0 o C to 126 g of water at 17.2 o C. The system comes to equilibrium at 19.1 o C. What is the metal?You add 14.2 g of a metal at 98.0 o C to 126 g of water at 17.2 o C. The system comes to equilibrium at 19.1 o C. What is the metal? q lostq lost m c  T  T T fT f T iT i  q gained  m  c   T  T f  T i 14.2 g ? -78.9 K 292.3 K 371.2 K 126 g 4.183 1.9 K 292.3 K 290.4 K

7. Work q = m c  Tq = m c  T Water sideWater side q = 126 g (4.183) 1.9K =q = 126 g (4.183) 1.9K = 1001.4102 J = -1001.4102 J lost1001.4102 J = -1001.4102 J lost -1001…= 14.2 g (c) -78.9 K-1001…= 14.2 g (c) -78.9 K c =.894 J/g Kc =.894 J/g K Most likely aluminum.Most likely aluminum. How close your answer is to the actual answer depends on how good your lab technique is.How close your answer is to the actual answer depends on how good your lab technique is.

8. Problems 2.28 mol of aluminum at 66.0 o C is dropped in water at 28.0 o C, the system comes to equilibrium (both temperatures are equal) at 34.0 o C, how much water was present?2.28 mol of aluminum at 66.0 o C is dropped in water at 28.0 o C, the system comes to equilibrium (both temperatures are equal) at 34.0 o C, how much water was present? q lost = 2.28 mol(24.2)(-32)q lost = 2.28 mol(24.2)(-32) = -1765.632 = -(n (75.3)(6.0))= -1765.632 = -(n (75.3)(6.0)) n = 3.9 mol H 2 On = 3.9 mol H 2 O

9. A different problem You add 37.1 g of a metal at 101.0 o C to 132 g of water at 19.5 o C. The system comes to equilibrium at 20.2 o C. What is the metal?You add 37.1 g of a metal at 101.0 o C to 132 g of water at 19.5 o C. The system comes to equilibrium at 20.2 o C. What is the metal? q lostq lost m c  T  T T fT f T iT i  q gained  m  c   T  T f  T i 37.1 g ? -80.8 K 293.4 K 374.2 K 132 g 4.183.7 K 293.4 K 292.7 K

10. Work q = m c  Tq = m c  T Water sideWater side q = 132 g (4.183).7K =q = 132 g (4.183).7K = 386.5092 J = -386.5092 J lost386.5092 J = -386.5092 J lost -386…= 37.1 g (c) -80.8 K-386…= 37.1 g (c) -80.8 K c =.129 J/g Kc =.129 J/g K Most likely lead.Most likely lead.

11. Another 5.25 mol of He at 34 o C is mixed with 24.3 mol of H 2 and the system comes to equilibrium at 14 o C, what was the initial temperature of the hydrogen?5.25 mol of He at 34 o C is mixed with 24.3 mol of H 2 and the system comes to equilibrium at 14 o C, what was the initial temperature of the hydrogen? q lost He = 5.25 mol(25.2)(287-307)q lost He = 5.25 mol(25.2)(287-307) q lost He = -2646 Jq lost He = -2646 J 2646 J = 24.3 mol(28.8) (287- T i H )2646 J = 24.3 mol(28.8) (287- T i H ) T i H2 = 283 K (10. o C)T i H2 = 283 K (10. o C)

12. Last One How many grams of silver at 50 o C would be required to heat 21 mol of water from 12 o C to 32 o C (bring the system at equilibrium at 32 o C)?How many grams of silver at 50 o C would be required to heat 21 mol of water from 12 o C to 32 o C (bring the system at equilibrium at 32 o C)? q = 21 mol(75.3)(20K) = 31626 Jq = 21 mol(75.3)(20K) = 31626 J q = - 31626 J = n (25.3) (-18)q = - 31626 J = n (25.3) (-18) n = 69.4 moln = 69.4 mol 69.4 x 107.9 g/mol = 7500 g or 7.5 kg69.4 x 107.9 g/mol = 7500 g or 7.5 kg